Science:Math Exam Resources/Courses/MATH103/April 2010/Question 08/Solution 1

We want to use the disk method and so we need to take the inverse of the function $\ f(x)=x(1-x).$

$f(x)=y=x-x^{2}\$

$x^{2}-x+y=0\$

$x_{1/2}={\frac {1\pm {\sqrt {1-4y}}}{2}}$

We get two possible inverse functions $f_{1}^{-1}(y)={\frac {1}{2}}+{\frac {1}{2}}{\sqrt {1-4y}}\$ and $f_{2}^{-1}(y)={\frac {1}{2}}-{\frac {1}{2}}{\sqrt {1-4y}}.\$

On the picture we see that for our purpose we need the function $f_{2}^{-1}$. To find the boundaries for the integral we calculate

$f(x=0)=0,\qquad f\left(x={\frac {1}{2}}\right)={\frac {1}{4}}.$

Now we use the formula for the disk method $V=\int \pi r(z)^{2}dz\$. Here $\ V$ is the volume and $\ r$ is the function of the radius. We need the function $f_{2}^{-1}$ for the radius.

${\begin{aligned}V&=\int _{0}^{\frac {1}{4}}\pi \left({\frac {1}{2}}-{\frac {\sqrt {1-4y}}{2}}\right)^{2}\,dy\\&=\pi \int _{0}^{\frac {1}{4}}\left({\frac {1}{4}}-{\frac {1}{2}}{\sqrt {1-4y}}+{\frac {1-4y}{4}}\right)\,dy\\&=\pi \int _{0}^{\frac {1}{4}}\left({\frac {1}{2}}-{\frac {1}{2}}{\sqrt {1-4y}}-y\right)\,dy\\&={\frac {\pi }{2}}\int _{0}^{\frac {1}{4}}\left(1-(1-4y)^{\frac {1}{2}}-2y\right)\,dy\\&={\frac {\pi }{2}}\left[y-{\frac {2}{3}}(1-4y)^{\frac {3}{2}}\left(-{\frac {1}{4}}\right)-y^{2}\right]_{0}^{\frac {1}{4}}\\&={\frac {\pi }{2}}\left({\frac {1}{4}}+{\frac {1}{6}}\left(1-4(1/4)\right)^{\frac {3}{2}}-(1/4)^{2}-0-{\frac {1}{6}}(1-4(0))^{\frac {3}{2}}+0^{2}\right)\\&={\frac {\pi }{2}}\left({\frac {1}{4}}-{\frac {1}{16}}-{\frac {1}{6}}\right)\\&={\frac {\pi }{96}}\end{aligned}}$